Physics 53
Gravity
Nature and Nature's law lay hid in night:
God said, "Let Newton be!" and all was light.
— Alexander Pope
Kepler’s laws
Explanations of the motion of the celestial bodies — sun, moon, planets and stars — are
among the oldest scientific theories. The apparent rotation of these bodies around the
earth every day was attributed by the ancients to their place in "celestial spheres" which
revolved daily around the stationary earth, assumed to be at the center of the universe.
The wandering of the planets against the background of stars presented a challenge to
this earth-centered (“geocentric”) model. Their peculiar motions were explained in
terms of smaller spheres rolling on larger ones. As observations became more accurate
these explanations became more intricately complicated.
The sun-centered (“heliocentric”) model of Copernicus (about 1540) was simpler, but it
made the earth merely one of the planets orbiting the sun. This provoked outrage, since
it challenged the central importance of homo sapiens in the universe.
In the late 1500’s the Danish astronomer Tycho Brahe carried out a careful and
systematic set of observations of the night sky, especially of the motions of the planets.
His assistant Johannes Kepler inherited the voluminous data he had accumulated;
Kepler subjected them to careful numerical analysis, and in the early years of the next
century he was able to discern in the data three regularities:
1. The orbits of planets are ellipses, with
the sun at one focus.
2. As the planet moves, the line from the
sun to a planet sweeps out equal areas
in equal times.
Kepler’s laws of planetary motion
3. The ratio T 2 / a3 is the same for all
planets, where T is the period and a is
the semi-major axis of the orbit.
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Gravity
How to explain these three laws (the "Kepler problem") was a major concern of the
scientists of the 17th century, among whom was Isaac Newton.
Universal gravitation
Newton's analysis of gravity was partly motivated by the Kepler problem, but more by
the idea that the moon in its orbit is continually “falling” toward the earth, much as
objects near the surface of the earth — including the famous apple? — fall to the
ground. He proposed that the solution to both problems lay in a universal law:
Every pair of point masses m1 and m2
attract with a force given by F = G
m1m2
,
r2
where r is the distance between the masses
and G is a universal constant.
Law of universal gravitation
Because of its simplicity, elegance and its universality — note that it applies to every pair
of point masses in the universe — this has served over the centuries as a model for what
a general law of nature ought to be like.
The value of G is 6.67 × 10−11 in SI units. This is very small, so the gravitational
attraction between everyday objects is too weak to be easily detectable — which is why
Newton never knew its value. Gravity is much weaker than the electric force between
two moderate sized charges. It is important in our everyday lives because:
•
The earth is very massive ( 5.98 × 10 24 kg). This means the attraction of the earth for
ordinary objects is a substantial force.
•
Atoms and molecules are electrically neutral, so ordinary objects do not usually
attract or repel each other electrically unless they get very close together.
Because mass is positive, the gravitational force is always an attraction.
Because electric charge can be either positive or negative, electrical forces can be either attractive or
repulsive, so they are subject to "shielding" which can reduce or negate their effects.
Gravity is “long range” in nature, meaning that its effects extend indefinitely far. In the
universe as a whole gravity is the most important force. In very massive stars it can
even overwhelm even the strong force that binds particles in the atomic nucleus.
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Gravity
Gravitational potential energy
Consider two masses, M and m. Let M be fixed at the origin and let the position vector
of m relative to the origin be r. In vector form, the gravitational force by M on m is
F(on m) = −G
Mm
r2
r̂ ,
where r̂ is a unit vector from M toward m. This shows that gravity is a central force, i.e.,
it is always directed toward the “force center” — in this case, M. Since all central forces
are known to be conservative, there is a potential energy function for this force, which
we can construct from the definition of potential energy:
⎛ Mm Mm ⎞
r
U(r) − U(r0 ) = − ∫ F dr = −G ⎜
−
.
r0
r0 ⎟⎠
⎝ r
We choose the potential energy to be zero when the masses are infinitely far apart (the
usual choice), so that as r0 → ∞ , U(r0 ) → 0. This gives us a relatively simple formula:
Gravitational potential energy (point
masses)
U = −G
Mm
r
Here r is the distance between the masses. For a system of more than two point masses,
each pair contributes to the total according to this formula.
Note that U is always negative, which is characteristic of an attractive force.
The gravitational field
The total gravitational force on a given particle due to a number of other particles can
be obtained (by superposition) from the formula for the force between pairs of particles.
In this way one could (in principle) calculate the force on a particular particle due to all
the particles in the earth.
But it is more convenient to regard the particles in the earth as a set of given "sources" of
gravity, and then to regard the force they exert on another particle as a collective effect.
This is what we do when we say that the earth as a whole exerts a force mg on a particle
of mass m near its surface. The quantity g represents the collective gravitational effect of
all the particles in the earth.
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Gravity
To describe the effect of the sources collectively is an essential idea in the concept of a
field. The sources “create” the field, and a “test” particle (the one we are interested in)
exhibits the effects of that field. But what is the field itself?
A field is a physical quantity that has a specific value at each point in space.
For gravity, we make the following definition:
The gravitational field g at a point in space is
defined as the gravitational force on a
particle of unit mass placed at that point.
Gravitational field
That is, if the gravitation force on a particle of mass m at a particular place is Fg , then
the gravitational field at that place is given by g = Fg /m . Conversely, if the field g is
known at a point in space, and if a particle of mass m is placed there, then it will
experience a gravitational force Fg = mg .
If gravity is the only force on the particle, then the value of the gravitational field — the
force per unit mass on the particle — is the same as the particle's acceleration. (The m on
both sides of F = ma cancels, so a = g.) The “acceleration of gravity” introduced earlier
for objects near the earth's surface is in fact just the gravitational field in that region.
As stated above, a field is a physical quantity that has a definite value at each point in space. The “value”
of the field at a point can be a scalar or a vector, so we have scalar fields and vector fields. We see that g is
a vector field. The temperature at each point in a room is an example of a scalar field. Other fields are the
velocity (vector) field in moving fluid, and the pressure (scalar) field in a fluid, to be discussed later in
this course. In the next course much time will be spent studying electric and magnetic fields, which are
vector fields, and electrostatic potential, which is a scalar field.
Why do we introduce the concept of a field? There are several reasons. Some are matters
of convenience, while others are of fundamental physical importance.
•
Knowledge of the field makes it relatively easy to understand what will happen
to any “test particle” placed at a point in the field.
•
For vector fields, one can make drawings (maps) of the field by constructing
curves ("field lines") tangent to the direction of the field. For scalar fields one can
draw curves connecting points where the field's value is the same. These maps
are valuable aids to the intuition.
•
By introducing fields we are able to avoid the untenable assumption of “action at
a distance” in which the interaction is assumed to be transmitted instantly from
one object to the other, even though they may be light years apart. (Experiment
in fact shows that no information can travel faster than the speed of light.)
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Gravity
•
The electromagnetic interaction (to be discussed in the next course) is most easily
expressed in terms of fields. The fundamental equations of electromagnetism are
given in terms of these fields.
•
In the electromagnetic case we know from experiment that the fields themselves
possess energy and momentum; indeed, proper use of the conservation laws
requires inclusion of these “mechanical” aspects of the fields as part of the
system's energy and momentum. Light itself consists entirely of energy and
momentum possessed by the electromagnetic field, traveling through space as a
wave in the field. (Because gravity is so weak, it is difficult to test experimentally
whether the gravitational field also has these properties, although all current
theories assume it does.)
The concept of a field has developed over the past two centuries into one of the most
important ideas in science. Modern theories take this to an extreme, postulating that all
matter (including things usually classed as “particles”) should be described in terms of
fields that obey the rules of relativity and quantum theory. In this world view we and
everything around us are just complicated combinations of various quantum fields.
Spherical shells
In applying his gravitation law to the sun and planets, Newton assumed that the
distance r between them should be taken to be the distance between their centers. Is that
only an approximation?
Newton was able to show — using his newly developed integral calculus — that it is
not an approximation, because the external gravitational effect of a uniform spherical
mass distribution is exactly the same as it would be if all the mass of the sphere were
concentrated at its center.
The essential step in proving this is to show that it is
true for a thin spherical shell, since any spherical
distribution can be considered to be a layered
combination of concentric shells. Consider the
gravitational field of a spherical shell, as shown. The
field at the location of a test mass m can be shown
(proof at the end of this section) to be:
R
r
m
M
M
⎧
⎪−G 2 r̂ for r > R
g(r) = ⎨
r
⎪
0 for r < R
⎩
This shows two remarkably simple properties:
•
At points outside the shell the field is the same as though all the mass of the shell
were concentrated at its center.
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Gravity
•
At points inside the shell the field is zero.
Since any spherically symmetric object (such as the sun) can be treated as a
superposition of concentric spherical shells — like layers of an onion — the total force
exerted by such an object on an external mass is the same as if the entire mass of the
sphere were at its center. This justifies Newton's treatment of the sun and planets as
mass points analyzing planetary motion.
From the formulas for the field it is easy to obtain the potential energy for the system of
a shell and a point test mass m:
⎧ Mm
⎪−G r for r > R
U(r) = ⎨
⎪−G Mm for r < R
⎩
R
Note carefully that although the field inside the shell is zero, the potential energy is not. We have taken
the potential energy to be zero at infinite distance, and that is the only place where it will be zero. The
potential energy inside the shell has the same value as at the shell’s surface.
These properties of shells are useful in dealing with any spherical distribution of mass.
Proof of formulas for g for a spherical shell.
To find the field at the field point P shown we first
find the contribution of an infinitesimal ring on the
shell, all points of which are at distance s from the
field point. Let the mass of the ring be dm and let the
line from the center of the shell to P define the x-axis.
By the symmetry the field at P has only an xcomponent, which is
dgx = −G
dm
s2
s
R
θ
α
P
r
M
cos α .
The area of the ring is its circumference times its width, or 2π R sin θ ⋅ Rdθ . It mass is
therefore
dm =
M
4π R
2
⋅ 2π R 2 sin θ dθ =
M
sin θ dθ .
2
Thus we have
dgx = −
GM cos α
sin θ dθ .
2 s2
The variables s, α, and θ are related by the geometry. From the law of cosines we have
R 2 = r 2 + s 2 − 2rscos α and s 2 = R 2 + r 2 − 2Rr cosθ . The first of these gives
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Gravity
s2 + r 2 − R 2
cos α =
,
2rs
while the second gives (since r and R are constants)
sds = Rr sin θ dθ .
Putting it all together we find
dgx = −
GM ⎛
r 2 − R2 ⎞
1
+
⎜
⎟ ds .
4Rr 2 ⎝
s2 ⎠
To find the field we integrate over s. For points outside the shell ( r > R ) we integrate
from s = r − R to s = r + R , and find
gx = −
GM
r2
for r > R .
For points inside the shell we integrate from s = R − r to s = R + r , and find
gx = 0 for r < R .
This proves the claimed properties.
Tidal forces
Besides solving the Kepler problem and relating the period of the moon to g at the
surface of the earth, Newton also was able to explain the ocean tides as a gravitational
effect caused by the moon and sun.
Shown is the earth, with the moon to the right
at distance r from the center of the earth. The
radius of the earth is R. Also shown are two
particles of mass m located at points 1 and 2
on opposite sides of the earth. If the moon
were absent, each of these masses would be
attracted toward the center of the earth by the
usual gravitational force mg .
R
2
1
to moon
We will analyze how the moon alters the forces on these two particles, as measured by
observers at points 1 and 2. These observers use a reference frame fixed on the earth,
which is a non-inertial frame because the earth is accelerating toward the moon.
We ignore (for the moment) the effects of the sun.
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Gravity
At the center of the earth let the gravitational field of the moon be g 0 , directed toward
the moon (to the right in the diagram) with magnitude
g0 =
GMM
r2
.
This is also the acceleration of the earth’s CM toward the moon, and it is therefore the
acceleration of the reference frame of our two observers.
At point 1 the gravitational field of the moon is g1 with magnitude
g1 =
GMM
(r − R)2
.
Similarly at point 2 the field of the moon is g 2 with magnitude
g2 =
GMM
(r + R)2
.
All of these vectors are directed toward the moon, of course.
But what the observers will measure is the effective gravity of the moon in their
reference frame, which we get by subtracting the reference frame’s acceleration g 0 .
eff
At point 1 the moon’s effective gravity is g1 = g1 − g 0 . Since g1 > g0 , this vector is
eff
toward the moon. The mass at that point experiences an outward force mg1 exerted by
the moon, away from the center of the earth. The total gravitational force on it (toward
eff
the center of the earth) is thus F1 = m(g − g1 ) .
eff
At point 2 we have g 2 = g 2 − g 0 . Since g2 < g0 , this vector is away from the moon, and
eff
therefore away from the center of the earth. The force of the moon on this mass is mg 2 .
eff
The total gravitational force on it (toward the center of the earth) is F2 = m(g − g2 ) .
Thus the moon produces forces on both of the particles away from the earth’s center,
reducing their gravitational attraction toward the earth’s center. If the particles are part
of the oceans there is a small outward bulge in the water surface at both points. This
gives rise to the twice daily tides. The rise or fall of the level in mid-ocean is tiny, but the
effect at shorelines of the resulting shift of large volumes of water toward or away from
the shore can be quite dramatic, especially if it is amplified by channel effects and wave
resonance.
Now we calculate the magnitude of these effects. We find
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Gravity
⎡ 1
1⎤
eff
g1 = g1 − g0 = GMM ⎢
−
⎥,
2
r2 ⎦
⎣ (r − R)
⎡1
⎤
1
eff
g2 = g0 − g2 = GMM ⎢ 2 −
⎥.
(r + R)2 ⎦
⎣r
These are not quite the same. But using the fact that R << r, we can make an
approximation. We write
1
1
1
=
⋅
(r ± R)2 r2 (1 ± R /r)2
and use the binomial approximation on the second factor:
(1 ± R /r)−2 ≈ 1  2R /r.
This gives
eff
eff
g1 ≈ g2 ≈
2GMM R
r3
.
Thus the effects on opposite sides of the earth are approximately equal and proportional
to the ratio MM /r 3 .
3
The sun also gives rise to tidal forces, but because of its much greater distance the ratio M / r for the sun
is only about half that for the moon. When earth, sun and moon are on the same line (with sun and moon
either on the same or opposite sides of the earth), their effects add and give rise to especially large tides,
called spring tides. When the lines from the sun and moon to the earth make a right angle, the effects tend
to cancel and we have small neap tides.
An indirect effect of tidal forces is to produce dissipative torques which reduce the
rotational kinetic energy of the object acted upon. This is why the moon (on which the
earth’s tidal forces are quite strong because the earth’s mass is so large) has long since
stopped rotating relative to the earth and keeps the same face toward us. Presumably
the moon is also making the earth slow in its daily rotation, but very gradually.
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Gravity